2C6H6(l)+15O2(g)-> 12CO2(g)+6H20(l)+6542 kj
If 6.200 g of C6H6 is burned and the heat produced from the burning is added to 5691 g of water at 21 °C, what is the final temperature of the water?
To find the final temperature of the water, we can use the concept of heat transfer. The heat released from burning benzene (C6H6) will be transferred to the water, causing a change in its temperature.
To calculate the final temperature, we need to use the formula:
q = m * c * ΔT
Where:
q is the heat transferred
m is the mass of the water
c is the specific heat capacity of water
ΔT is the change in temperature
First, let's calculate the heat transferred to the water. We know that the heat released from burning benzene is 6542 kJ, so:
q = 6542 kJ
Next, let's calculate the mass of water in grams:
mass of water = 5691 g
Now, we need to determine the specific heat capacity of water. The specific heat capacity of water is 4.18 J/(g°C) or 1 cal/(g°C).
Using the specific heat capacity value in Joules per gram and Celsius, let's convert the heat transferred to Joules:
q = 6542 kJ * 1000 J/1 kJ
q = 6,542,000 J
Now, we can substitute the values into the formula:
6,542,000 J = (5691 g) * (4.18 J/(g°C)) * ΔT
Let's solve for ΔT:
ΔT = 6,542,000 J / (5691 g * 4.18 J/(g°C))
ΔT ≈ 290 °C
Therefore, the final temperature of the water would be 290 °C.